University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 16 - Practice Exercises - Page 16-34: 15

Answer

$xy=-y^3+c$

Work Step by Step

The standard form of the given equation is: $x dy+y dx=-3y^2 dy$ ....(1) In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides. $\int [xy]' =\int -3y^2 dy$ Thus, we have $xy=-y^3+c$
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